Leetcode 2. Add Two Numbers

https://leetcode.com/problems/add-two-numbers/

 

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
  ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
      /*
      可能长度+1,l1为返回答案.[假设l1,l2结构不需要了]
      */
      ListNode *ll1=l1,*ll2=l2;
      int a=ll1->val,b=ll2->val;
      ll1->val=(a+b)%10;
      int carry_bit=(a+b)/10;//进位
      while(ll1->next && ll2->next){
          /*
          10进制加法,直到某一个列表到末尾,相加后结束.
          */
          a=ll1->next->val,b=ll2->next->val;
          ll1->next->val=(a+b+carry_bit)%10;
          carry_bit=(a+b+carry_bit)/10;
          ll1=ll1->next;
          ListNode *tmp=ll2;
          ll2=ll2->next;
          delete tmp;
      }
      if(ll2->next) ll1->next=ll2->next;//ll1设为长链
      while(ll1->next && carry_bit){
          /*
          进位的处理
          */

          a=ll1->next->val;
          ll1->next->val=(a+carry_bit)%10;
          carry_bit=(a+carry_bit)/10;
          ll1=ll1->next;
      }
      if(carry_bit==1){
          /*
          长度+1的情况
          */
          ListNode* p=new ListNode(1);
          ll1->next=p;
      }
      return l1;
  }
};

 python版本

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        p=l1
        e=0
        x=l1.val+l2.val+e
        e=x//10
        l1.val=x%10
        while l1.next and l2.next:
            x=l1.next.val+l2.next.val+e
            e=x//10
            l1.next.val=x%10
            l1=l1.next
            l2=l2.next
        if l2.next:
            l1.next=l2.next
        while l1.next:
            x=l1.next.val+e
            e=x//10
            l1.next.val=x%10
            l1=l1.next
        if e!=0:
            x=ListNode(e)
            l1.next=x
        return p